Question

A metal piece weighs 200 gf in air and 150 gf when completely immersed in water. (i) Calculate the relative density of the metal piece. (ii) How much will it weigh in a liquid of density 0.8 gcm-3?. please explain with formulas​

Answer 1

As we know that relative density =density of material/density of water
And density =m/v
So we relation that RD=m1/m2
M1=200
M2=150
Rd=200/150=4/3

Answer 2

Thus the wight in a liquid of density 0.8 gcm^-3 is 199.8 g

Explanation:

• m = Mass of solid
• V = Volume of solid
• ρs= Density of solid
• ρw= Density of water
• g = Acceleration due to gravity
• Fb = Bouyant force (= V ρwg)

In air

200 gf = mg = Vρsg ...[∵Mass = Volume × Density]

Let’s call this as equation (1)

In water

150 gf = mg - Fb

150 gf = 200 gf – Fb

Fb=50 gf

Vρwg = 50 gf ………(2)

Divide equations (1) and (2)

Vρsg / Vρwg = 200 gf / 50 gf = 4

Relative density is

RD = Density of solid/Density of water = ρs/ρw = 10

First we use equation no 2 to find V;

V(pwg) = 50 gf

Density of water is 1000 kg/m^3;

V = 50/1000 f

=   1/20 f

Now calculating weight in another liquid of this solid;

Weight = 200 gf – Vpg

Weight = 200 gf – 1/20 f (0.8 g/3) g

           = 200 gf – 8/60fg

           = 199.8 g

Thus the wight in a liquid of density 0.8 gcm^-3 is 199.8 g